Ćw.409 : Wyznaczanie stężenia roztworu za pomocą kalorymetru
fotoelektrycznego.

i

1 11

..

:=

l

1 10

..

:=

k

1 3

..

:=

m

10

:=

n

1 m

..

:=

TABELA I

λ

i

480

490

500

510

520

530

540

550

560

570

580

:= E

i

0.6

0.65

0.73

0.79

0.75

0.65

0.48

0.33

0.215

0.14

0.095

:=

480

490

500

510

520

530

540

550

560

570

580

0

0.2

0.4

0.6

0.8

E

i

λ

i

TABELA II

c

l

45

40

35

30

25

20

15

10

5

:= E1

l

0.055

0.05

0.04

0.035

0.025

0.01

0.005

0.004

0.003

:= E2

l

0.06

0.055

0.04

0.03

0.02

0.01

0.006

0.004

0.004

:= E3

l

0.06

0.055

0.04

0.03

0.02

0.01

0.006

0.004

0.003

:= Esr

l

0.538

0.497

0.408

0.343

0.302

0.223

0.180

0.097

0.062

:=

Esr

l

E1

l

E2

l

+

E3

l

+

3

:=

a

slope c Esr

,

(

)

:=

b

intercept c Esr

,

(

)

:=

a

1.319

10

3

−

×

=

b

6.257

−

10

3

−

×

=

y

l

a c

l

⋅

b

+

:=

0

4.5

9

13.5

18

22.5

27

31.5

36

40.5

45

0.01

−

3

−

10

3

−

×

4 10

3

−

×

0.011

0.018

0.025

0.032

0.039

0.046

0.053

0.06

KRZYWA

Esr

l

y

l

c

l

1) Wyznaczenie stężenia
nieznanego roztworu:

2)Dyskusja błędów :

Ex

k

0.025

0.026

0.025

:=

∆a

m

m

2

+

1

m

n

Esr

n

( )

2

∑

=

a

1

m

n

c

n

Esr

n

⋅

(

)

∑

=

⋅

−

b

1

m

n

Esr

n

∑

=

⋅

−

m

1

m

n

c

n

( )

2

∑

=

⋅

1

m

n

c

n

∑

=

















2

−

⋅

:=

∆b

1

m

n

c

n

( )

2

∑

=

1

m

⋅

∆a

⋅

:=

Ex

sr

k

Ex

k

∑

3

:=

c

x

Ex

sr

b

−

a

:=

Ex

sr

0.025

=

c

x

23.946

=

∆a

1.21

10

4

−

×

=

∆b

3.23

10

3

−

×

=